Graph Operations

Introduction
Depth-First Search

Uses for Depth-First Search

Test Yourself #1

Breadth-First Search
Summary
Answers to Self-Study Questions

Introduction

As discussed in the introduction to graphs notes, graphs are often a good representation for problems involving objects and their relationships because there are standard graph operations that can be used to answer useful questions about those relationships. Here we discuss two such operations: depth-first search and breadth-first search, and some of their applications.

Both depth-first and breadth-first search are "orderly" ways to traverse the nodes and edges of a graph that are reachable from some starting node. The main difference between depth-first and breadth-first search is the order in which nodes are visited. Of course, since in general not all nodes are reachable from all other nodes, the choice of the starting node determines which nodes and edges will be traversed (either by depth-first or breadth-first search).

Depth-first Search

Depth-first search can be used to answer many questions about a graph:

is it connected?
is there a path from node j to node k?
does it contain a cycle?
what nodes are reachable from node j?
can the nodes be ordered so that for every node j, j comes before all of its successors in the ordering?

The basic idea of a depth-first search is to start at some node n, and then to follow an edge out of n, then another edge out, etc, getting as far away from n as possible before visiting any more of n's neighbors. To prevent infinite loops in graphs with cycles, we must keep track of which nodes have been visited. Here is the basic algorithm for a depth-first seach from node n:

mark n "visited"
recursively do a depth-first search from each of n's unvisited successors

Information about which nodes have been visited can be kept in the nodes themselves (e.g., using a boolean field) or in an auxiliary array of booleans of size N (where N is the number of nodes in the graph). In both cases, all nodes should be initialized to "unvisited". Below is code for depth-first search, assuming that visited information is in a node field named "visited", and that each node's successors are in a Sequence field named "successors". Note that this basic depth-first search doesn't actually do anything except mark nodes as having been visited. We'll see in the next section how to use variations on this code to do useful things.

static void dfs (GraphNode n) {
   n.visited = true;
   Sequence S = n.successors;
   for (S.start(); S.isCurrent(); S.advance()) {
      GraphNode m = S.getCurrent();
      if (! m.visited) dfs(m);
   }
}

Here's a picture that illustrates the dfs method. In this example, node numbers are used to denote the nodes themselves (i.e., the call dfs(0) really means that the dfs method is called with a pointer to the node labeled 0). Two different colors are used to indicate the node currently being visited and the previously visited node.

Note that in the example illustrated above, the order in which the nodes are visited is: 0, 2, 3, 1, 4. Another possible order (if node 4 were the first successor of node 0) is: 0, 4, 2, 3, 1.

To analyze the time required for depth-first search, note that one call is made to dfs for each node that is reachable from the start node. Each call looks at all successors of the current node, so the time is O(# reachable nodes + total # of outgoing edges from those nodes). In the worst case, this is all nodes and all edges, so the worst-case time is O(N + E).

Uses for Depth-First Search

Recall that at the beginning of this section we said that depth-first search can be used to answers questions about a graph such as:

is it connected?
is there a path from node j to node k?
does it contain a cycle?
what nodes are reachable from node j?
can the nodes be ordered so that for every node j, j comes before all of its successors in the ordering?

Questions 2, 3 and 5 are discussed; the others are left as exercises.

Path Detection

The first question we will consider is: is there a path from node j to node k? This question might be useful, for example:

When the graph represents airline routes, and we want to ask "Can I fly from Madison to London (maybe w/ some connections)?", or
when the graph represent CS course prerequisites, and we want to ask "Is CS 367 a (transitive) prerequisite for CS 640?"

To answer the question, do the following:

step 1: mark all nodes "not visited"
step 2: dfs(j)
step 3: there is a path from j to k iff k is marked "visited"

Cycle Detection

There are two variations that might be interesting:

does a graph contain a cycle?
is there a cyclic path starting from node j?

Consider the example given above to illustrate depth-first search. There is a cycle in that graph starting from node 0. Is there something that happens during the depth-first search that indicates the presence of that cycle?? Note that during dfs(1), 0 is a successor of 1, but is already visited. But that isn't quite enough to say that there's a cycle, because during dfs(3), node 4 is a successor of 3 that has already been visited, but there is no cycle starting from node 4.

What's the difference? The answer is that when node 0 is considered as a successor of node 1, the call dfs(0) is still "active" (i.e., its activation record is still on the stack); however, when node 4 is considered as a successor of node 3, the call dfs(4) has already finished. How can we tell the difference?? The answer is to keep track of when a node is "inProgress" (as well as whether it has been visited or not). We can do this by using a "mark" field with three possible values:

unvisited
inProgress
done

instead of the boolean "visited" field we've been using. Initially, all nodes are marked "unvisited". When the dfs method is first called for node n, it is marked "inProgress". Once all of its successors have been processed, it is marked "done". There is a cyclic path reachable from node n iff some node's successor is found to be marked "inProgress" during dfs(n).

Here's the code for cycle detection:

static boolean hasCycle(GraphNode n) {
   n.mark = inProgress;
   Sequence S = n.successors;
   for (S.start(); S.isCurrent(); S.advance()) {
      GraphNode m = S.getCurrent();
      if (m.mark == inProgress) return true;
      if (m.mark != done) {
         if (hasCycle(m)) return true;
      }
   }
   n.mark = done;
   return false;
}

Note that if we want to know whether a graph contains a cycle anywhere (not just one that is reachable from node n) we might have to call hasCycle at the "top-level" more than once:

static boolean graphHasCycle(Graph G) {
    mark all nodes unvisited;
    for each node k in the graph {
       if (node k is marked unvisited) {
          if (hasCycle(k)) return true;
       }
    }
    return false;
}

Topological Numbering

Think again about the graph that represents course prerequisites. As long as there are no cycles in the graph (which wouldn't make sense, because it would mean that a course was a prerequisite for itself!) there is at least one order in which to take courses, such that all prereqs are satisfied; i.e., so that for every course, all prerequisites are taken before the course itself is taken.

Topological numbering addresses exactly this problem. The goal is to assign numbers to nodes so that for every edge j -> k, the number assigned to j is less than the number assigned to k. A topological numbering of the prerequisites graph would tell you one legal order in which to take the CS courses. For example:

To find a topological numbering, we use a variation of depth-first search. The intuition is as follows:

As long as there are no cycles in the graph, there must be at least one node with no outgoing edges.
The last number (N) can be given to any such node (310, 577, or 640 in our example).
Once all of a node's successors have numbers, the node itself can get the next smallest number.

These 2 situations correspond to the point in method hasCycle where node n is marked "done" (when it has no more unvisited successors). We just need to keep track of the current number. Below is a method that, given a node n and a number num, assigns topological numbers to all unvisited nodes reachable from n, starting with num and working down. Note that before calling this method for the first time, all nodes should be marked "unvisited", and that the initial call should pass N (the number of nodes in the graph) as the 2nd parameter.

static int topNum (GraphNode n, int num) throws CycleException {
    n.mark = inProgress;
    Sequence S = n.successors;
    for (S.start(); S.isCurrent(); S.advance()) {
       GraphNode m = S.getCurrent();
       if (m.mark == inProgress) {
           // no topological ordering for a cyclic graph!
           throw new CycleException();
       }
       if (m.mark != done) num = topNum(m, num);
    }
    // here when n has no more successors
    n.mark = done;
    n.number = num;
    return num-1;
}

As was the case for cycle detection, we might need several "top-level" calls to number all nodes in a graph.

TEST YOURSELF #1

Question 1: Give two different topological numberings for the following graph.

Question 2: The topNum method given above only assigns numbers to the nodes reachable from node n. Write psuedo code for method numberGraph, similar to the code given for method graphHasCycle above, that assigns topological numbers to all nodes in a graph. Assume that a Graph has a numNodes method that returns the number of nodes in the graph.

Question 3: Write a method isConnected, that returns true iff its Graph parameter is connected. Assume that every node has a list of its predecessors as well as a list of its successors.

solution

Breadth-first Search

Breadth-first search provides another "orderly" way to visit (part of) a graph. The basic idea is to visit all nodes at the same distance from the start node before visiting farther-away nodes. Like depth-first search, breadth-first search can be used to find all nodes reachable from the start node. It can also be used to find the shortest path between two nodes in an unweighted graph.

Breadth-first search uses a queue rather than recursion (which actually uses a stack); the queue holds "nodes to be visited". If the graph is a tree, breadth-first search gives you a level-order traversal. Here's the pseudo code:

static void bfs (GraphNode n) {
  Queue Q = new Queue();
  GraphNode current;

  n.visited = true;
  Q.enqueue(n);
  while (! Q.isEmpty())){
     current = Q.dequeue();
     Sequence S = current.successors;
     for (S.start(); S.isCurrent(); S.advance()) {
        GraphNode k = S.getCurrent();
        if (! k.visited){
            k.visited = true;
            Q.enqueue (k);
        } // end if k not visited
     } // end for every successor k
  } // end while Q not empty
}

Here's the same example graph we used for depth-first search:

The order in which nodes are "visited" as a result of bfs(0) is:

As with depth-first search, all nodes marked "visited" are reachable from the start node, but nodes are visited in a different order then they would be using depth-first search.

We can use a variation of bfs to find the shortest distance (the length of the shortest path) to each reachable node:

keep an array of distances; fill in distance[n] = 0
when a node k is about to be enqueued, set distance[k] = distance[current] + 1

This technique only works in unweighted graphs (i.e., in graphs in which all edges are assumed to have length 1). An interesting problem is how to find shortest paths in a weighted graph; i.e., given a "start" node n, to find, for each other node m, the path from n to m for which the sum of the weights on the edges is minimal (assuming that no edge has a negative weight). For example, in the following graph, nodes represent cities, edges represent highways, and the weights on the edges represent distances (the length of the highway between the two cities). Breadth-first search can only tell you which route from Madison to Green Bay goes through the fewest other cities; it cannot tell you which route is the shortest.

A clever algorithm that can be used to solve this problem (to find shortest paths in a weighted graph with non-negative edge weights) has been defined by Edsgar Dijkstra (and so is called "Dijkstra's algorithm"). The worst-case running time of the algorithm is O(E log N), assuming that edges are represented using adjacency lists, not an adjacency matrix, where E is the number of edges and N is the number of nodes. You can find a description of the algorithm in our textbook.

SUMMARY

A graph a set of nodes and a set of edges.
There are two kinds of graphs: directed and undirected.
Low-level operations include:

add a node
add an edge
remove a node
remove an edge
determine whether there is an edge j->k
find all successors / predecessors / neighbors of node j

High-level operations include:

depth-first search, which can be done on the entire graph (e.g., to find cycles or to produce a topological ordering), or on part of a graph (e.g., to determine which nodes are reachable from a given node)
breadth-first search, which can also be used to determine reachability, and can be used to find shortest paths in unweighted graphs
Dijkstra's algorithm, which finds shortest paths in weighted graphs